Sugestions with code

[jblackann]

You might want to take a look at this thread also.

http://picbasic.co.uk/forum/showthread.php?t=4289

[ruijc]

jblackann ,

thanks for the link.

But i'm getting a hard time understanding the logic.

AD_AN3_Volt = (AD_AN3_VALUE */ 500) >> 2

Using 10Bit resolution:

5V = 1023 so 2.5V = 511

( final reading on LCD should be 2.50V )

If i have 2.5V the AD_AN3_VALUE = 511

*/ -> get the middle 16bits from the 32 ( divided aftermath )

511 / 500 = 1,022

1,022 - the 16 middle bits ???

Then >>2 means shift 2 positions to the right ( binary ) ??


And final the DEC2 - this is not included in the PBP manual...
EDIT: But this last line i understand the logic.

.

[ruijc]

Can someone help me undestand the code ?

AD_AN3_Volt = (AD_AN3_VALUE */ 500) >> 2

thanks

[jblackann]

I had a problem understanding this too. Here is another link that you can look at that explains the '*/' operator. http://www.picbasic.co.uk/forum/showthread.php?t=5545

Quote from Darrel Taylor on Jan 17, 2007.
======================
*/ is the MidWord multiplier.

When you multiply 2 16-bit values together, you'll get up to a 32-bit result (4 bytes). The Midword, is the 2 bytes in the middle.

For instance, let's say you have an A/D value of 600, and you multiply it times 5000, that gives you 3,000,000. And if you express that in Hexadecimal, it looks like ...
002D:C6C0

The MidWord is 2DC6, or 11,718.

That's the same as (600 * 5000) / 256. The /256 is implied, and happens without the extra time required to do a division. So, it can be allot faster than using DIV32.

The rest of the formula is, >> 2, or shiftright 2 places. Which is the same as /4. Since the result has already been divided by 256, /4 makes the total divisor equal to 1024 (256*4). And, the result will be 2,929, which represents 2.929 Volts input to the ADC.

So, the whole formula is really just (600 * 5000) / 1024. But in a way that PBP can handle it with 16-bit numbers.
--------------------------------------------------------------------------------

OK, so how to go the other direction...

Quote:
I would like to take 0.01 to 0.250V

You'll need the integer numbers (10 to 250) I'll call it Offset. Then just do the reverse formula.

AD_Cal = (Offset * 1024) / 5000

This one doesn't fit in with the */ operator, so you'll need to use DIV32.
Code:
Offset = 250 ; 0.250V - Must be a WORD Var
AD_Cal = Offset * 1024
AD_Cal = DIV32 5000This gives a result of 51, which can then be added or subtracted from the AD result as necessary.

HTH,
__________________
DT
=====================

As for the DEC3 command, I looked in the manual at SEROUT2 command, it has a little bit more explanation. From the manual, a numerical value preceded by DEC will send the ASCII representation of its decimal value. For example, if B0 = 123, then DEC B0 (or DEC 123) will send "123".

Also BIN, DEC, and HEX may also be followed by a number. Normally these modifiers will display exactly as many digits as necessary, zero blanked (leading zeros are not sent). However, if a number follows the modifier, SEROUT2 will always send that number of digits, adding leading zeros as necessary. It was also trim off any extra high order digits. For exampl, BIN6 8 would send "001000" and BIN2 8 would send "00".

You can try to do some experimenting with this.

I believe with the code I supplied
======
TO have precision to three decimal points use:
AD_AN3_Volt = (AD_AN3_VALUE */ 5000) >> 2
LCDOUT $FE, 1 ' Clear LCD screen
LCDOUT "A to D Value"
LCDOUT $FE, $C0, DEC (AD_AN3_volt/1000) ,".",DEC3 AD_AN3_VOLT ,"VDC"
=======
I am using 10 bit A to D, so the max value I could have for AD_AN3_VOLT is 1023. I then use the */ 5000 which is equivalent to *5000 /256 and then I use '>>' (shift right by 2) which is equivalent to dividing by 4. My value is now equal to a max of 4995. I then divide AD_AN3_VOLT by 1000 and get a integer (using the DEC will allow it to display 0 if it needs to). And DEC3 AD_AN3_VOLT will provide a 3 digit value for AD_AN3_VOLT. I believe that I how it works. I did this awhile ago and proved it to myself then. I would suggest you experiment with the code and see that happens. Best of luck.

[ruijc]

jblackann,

i thank you very much for the detailed explanation

In fact i did tryed the code and it works 5*****

i just wasnt undestanding the math...and now i do

This way we get precise results with less code.


Thanks

.